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How do you express #cos(4theta)# in terms of #cos(2theta)#?

Answer 1

Madelyn Adams

#cos(4theta) = 2(cos(2theta))^2-1#

Start by replacing #4theta# with #2theta+2theta#

#cos(4theta) = cos(2theta+2theta)#

Knowing that #cos(a+b) = cos(a)cos(b)-sin(a)sin(b)# then

#cos(2theta+2theta) = (cos(2theta))^2-(sin(2theta))^2#

Knowing that #(cos(x))^2+(sin(x))^2 = 1# then

#(sin(x))^2 = 1-(cos(x))^2#

#rarr cos(4theta) = (cos(2theta))^2-(1-(cos(2theta))^2)#

# = 2(cos(2theta))^2-1#

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Answer 2

Violet Anderson

To express $\cos(4\theta)$ in terms of $\cos(2\theta)$ , you can use the double angle identities for cosine:

$\cos(2\theta) = 2\cos^2(\theta) - 1$

Using this identity, you can rewrite $\cos(4\theta)$ in terms of $\cos(2\theta)$ as follows:

$\cos(4\theta) = \cos(2(2\theta))$

$= 2\cos^2(2\theta) - 1$

$= 2(2\cos^2(\theta) - 1)^2 - 1$

$= 2(4\cos^4(\theta) - 4\cos^2(\theta) + 1) - 1$

$= 8\cos^4(\theta) - 8\cos^2(\theta) + 1$

Therefore, $\cos(4\theta) = 8\cos^4(\theta) - 8\cos^2(\theta) + 1$ in terms of $\cos(2\theta)$ .

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Answer from HIX Tutor

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